What Happens to Voltage When a Dielectric is Inserted into a Charged Capacitor?

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Inserting a dielectric into a charged capacitor affects the voltage across its plates. For a parallel plate capacitor with a dielectric constant of k=4, the new voltage can be calculated using the formula V=Q/C. The initial voltage was 12V, and after disconnecting from the battery, the charge remains constant while the capacitance increases due to the dielectric. The user calculated V as 3V but questioned its accuracy, suggesting it should be adjusted for the dielectric constant. Clarification on the complete question is needed to resolve the error in the voltage calculation.
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A very large parallel plate capacitor has two plates, each with an area of 9m^2, and a separation of 3nm between them.

If the capacitor was then disconnected from the battery (12V), and then a dielectric of k=4 inserted between the plates, fnd the new values for the following (after system has reached equilibrium)

C= 0.1062 F
Q= .3186 C
V= ____ V

I did V=Q/C = 3V, but I got it wrong. I don't see what I did wrong really.
 
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Take the dielectric constant of k=4 into consideration.
 
so it would be 3v/4? and that would give me the answer?
 
It would help if you could post the exact text of the question. You've left some parts out, and that is probably where the error happens. Can you please post the whole question verbatim?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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